The circuit shown in the figure below is connected for 3.3 min. (Assume R1 = 6.8

Question

The circuit shown in the figure below is connected for 3.3 min. (Assume R1 = 6.80ohms, R2 = 1.2ohms, and V = 14.0 V.)

(a) Determine the current in each branch of the circuit.

branch: magnitude (A) direction
left branch: ?
middle branch: ?
right branch: ?

(b) Find the energy delivered by each battery.
4.00 V battery: ? J
14.0 V battery: ? kJ

(c) Find the energy delivered to each resistor.
6.80 ohms resistor: ? J
5.00 ohms resistor: ? J
1.00 ohms resistor: ? J
3.00 ohms resistor: ? J
1.2 ohms resistor: ? J

(e) Find the total amount of energy transformed into internal energy in the resistors.
kJ

Firefox File Edit View History Bookmarks Tools Window Help C The circuit shown in The Fia x Ch. 28 Direct-current circ... x d https /www. webassign.net/ Assignment-Responsesllast?dep 15990369 weblStudent Search 6. -15 points serPSET9 28, P023 MI The circuit shown in the figure below is connected for 3.30 min. (Assume R 6.80 n, R 1.20 n, and V 14.0 V 5.00 (2 3.00 1.00 S2 4.00 V a) Determine the current in each branch of the circuit. branch magnitude (A) direction left branch Select-- Select-- middle branch right branch Select (b) Find the energy delivered by each battery. 4.00 V battery 14.0 V battery (c) Find the energy delivered to each resistor. 6.80 n resistor 5.00 n resistor 1.00 resistor 3.00 n resistor 1.20 n resistor (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. 45% Mon 9:57 AM My Notes Ask Your Teacher

Explanation / Answer

1)We name the current in the left segment, middle segment and right segment I1,I2,I3 respectively

From junction rule,

I1+I2= I 3----->1

Using loop rule for the loop containing the left and right segments alone we get,

(6.8 I1 + (1.20+3)I2 )V= 14 V----->2

Using the same rule for the loop containing the left and middle loop,

(6.8 I1 - (5+1)I2)V = 4V------>3

Substituting 2 in 1

I1 =( 14 - (4.2(I2)))/11------->4

Solving 2 for I2

I2= (4+ 6.8I1)/6----->5

Substitute 5 in 4

I1= (14-(0.7×(4+6.8 I1)))/11

11(I1)= 14-2.8-4.76(I1)

I1= 0.7 A

Then from 5,

I2 = 1.46 A

I3= 2.16 A

2) the energy delivered by each battery = VIt

For 4 V battery E = (4× 0.7 A× 3.3×60s) =554.4 J

For 14V battery E =(15× 2.16A×3.3×60s)= 6.52 KJ

3)E = I2Rt

for R = 6.8 ohm

E = (0.7A)2x(6.8)×(3.3× 60s) =659.74 J

For R=5 ohm

E =(1.46)2(5)(3.3× 60)=2.11 KJ

For R=1 ohm

E = (1.46)2(1)(3.3× 60)=422J

For R=3 ohms

E = ( 2.16)2(3)(198)=2.77KJ

For R=1.2 ohm

E= (2.16)2(1.2)(198)=1.11KJ

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